MHTCET 15 Apr 2026 Shift 2 Paper

Concept:In simple harmonic motion, the restoring force is directly proportional to the displacement from the mean position, given by $F = kx$.Explanation:The restoring force experienced by the body is $F = 0.4 \, \mathrm{N}$.The displacement from the mean position is $x = 4 \, \mathrm{cm}$, which is converted to metres as $x = 0.04 \, \mathrm{m}$.Using the relation $F = kx$, the force constant is calculated as:$$k = \frac{F}{x} = \frac{0.4}{0.04} = 10 \, \mathrm{N/m}$$Hence, the force constant of the oscillator is $10 \, \mathrm{N/m}$.To find the magnitude of acceleration, apply Newton's second law, $F = ma$.Rearranging the equation gives $a = \frac{F}{m}$.Substituting the given values $F = 0.4 \, \mathrm{N}$ and $m = 0.4 \, \mathrm{kg}$:$$a = \frac{0.4}{0.4} = 1 \, \mathrm{m/s^2}$$Therefore, the magnitude of acceleration is $1 \, \mathrm{m/s^2}$.Answer:The force constant is $10 \, \mathrm{N/m}$ and the magnitude of acceleration is $1 \, \mathrm{m/s^2}$.Hence, the correct option is Option B.